Problem: Baby Ehab was toying around with arrays. He has an array a of length n. He defines an array to be good if there's no way to partition it into 2 subsequences such that the sum of the elements in the first is equal to the sum of the elements in the second. Now he wants to remove the minimum number of elements in a so that it becomes a good array. Can you help him?

A sequence b is a subsequence of an array a if b can be obtained from a by deleting some (possibly zero or all) elements. A partitioning of an array is a way to divide it into 2 subsequences such that every element belongs to exactly one subsequence, so you must use all the elements, and you can't share any elements.

Input

The first line contains an integer n (2 ≤ n ≤ 100) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_{n} (1 ≤ a_i ≤ 2000) — the elements of the array a.

Output

The first line should contain the minimum number of elements you need to remove.

The second line should contain the indices of the elements you're removing, separated by spaces.

We can show that an answer always exists. If there are multiple solutions, you can print any.

Examples

Input


4
6 3 9 12


Output


1
2

Input


2
1 2


Output


0

Note

In the first example, you can partition the array into [6,9] and [3,12], so you must remove at least 1 element. Removing 3 is sufficient.

In the second example, the array is already good, so you don't need to remove any elements.

Solution (in PYTHON3):

Solution:
import sys
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
import bisect,string,math,time,functools,random,fractions
from bisect import*
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
rep=range;R=range
def I():return int(input())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def AI():return map(int,open(0).read().split())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
def RI(a=1,b=10):return random.randint(a,b)
def INP():
    N=10
    n=random.randint(1,N)
    a=RLI(n,0,n-1)
    return n,a
def Rtest(T):
    case,err=0,0
    for i in range(T):
        inp=INP()
        a1=naive(*inp)
        a2=solve(*inp)
        if a1!=a2:
            print(inp)
            print('naive',a1)
            print('solve',a2)
            err+=1
        case+=1
    print('Tested',case,'case with',err,'errors')
def GI(V,E,ls=None,Directed=False,index=1):
    org_inp=[];g=[[] for i in range(V)]
    FromStdin=True if ls==None else False
    for i in range(E):
        if FromStdin:
            inp=LI()
            org_inp.append(inp)
        else:
            inp=ls[i]
        if len(inp)==2:a,b=inp;c=1
        else:a,b,c=inp
        if index==1:a-=1;b-=1
        aa=(a,c);bb=(b,c);g[a].append(bb)
        if not Directed:g[b].append(aa)
    return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
    #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
    mp=[boundary]*(w+2);found={}
    for i in R(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
    mp+=[boundary]*(w+2)
    return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def accum(ls):
    rt=[0]
    for i in ls:rt+=[rt[-1]+i]
    return rt
def bit_combination(n,base=2):
    rt=[]
    for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
    return rt
def gcd(x,y):
    if y==0:return x
    if x%y==0:return y
    while x%y!=0:x,y=y,x%y
    return y
def YN(x):print(['NO','YES'][x])
def Yn(x):print(['No','Yes'][x])
def show(*inp,end='\n'):
    if show_flg:print(*inp,end=end)

mo=10**9+7
#mo=998244353
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
alp=[chr(ord('a')+i)for i in range(26)]
#sys.setrecursionlimit(10**7)
def gcj(c,x):
    print("Case #{0}:".format(c+1),x)

########################################################################################################################################################################
# Verified by
# https://yukicoder.me/problems/no/979
# https://atcoder.jp/contests/abc152/tasks/abc152_e

## return prime factors of N as dictionary {prime p:power of p}
## within 2 sec for N = 2*10**20+7
def primeFactor(N):
    i,n=2,N
    ret={}
    d,sq=2,99
    while i<=sq:
        k=0
        while n%i==0:
            n,k,ret[i]=n//i,k+1,k+1
        if k>0 or i==97:
            sq=int(n**(1/2)+0.5)
        if i<4:
            i=i*2-1
        else:
            i,d=i+d,d^6
    if n>1:
        ret[n]=1
    return ret

## return divisors of n as list
def divisor(n):
    div=[1]
    for i,j in primeFactor(n).items():
        div=[(i**k)*d for d in div for k in range(j+1)]
    return div

## return the list of prime numbers in [2,N], using eratosthenes sieve
## around 800 ms for N = 10**6  by PyPy3 (7.3.0) @ AtCoder
def PrimeNumSet(N):
    M=int(N**0.5)
    seachList=[i for i in range(2,N+1)]
    primes=[]
    while seachList:
        if seachList[0]>M:
            break
        primes.append(seachList[0])
        tmp=seachList[0]
        seachList=[i for i in seachList if i%tmp!=0]
    return primes+seachList


## retrun LCM of numbers in list b
## within 2sec for no of B = 10*5  and  Bi < 10**6
def LCM(b,mo=10**9+7):
    prs=PrimeNumSet(max(b))
    M=dict(zip(prs,[0]*len(prs)))
    for i in b:
        dc=primeFactor(i)
        for j,k in dc.items():
            M[j]=max(M[j],k)
    
    r=1
    for j,k in M.items():
        if k!=0:
            r*=pow(j,k,mo)
            r%=mo
    return r

## return (a,b,gcd(x,y)) s.t. a*x+b*y=gcd(x,y)
def extgcd(x,y):
    if y==0:
        return 1,0
    r0,r1,s0,s1 = x,y,1,0
    while r1!= 0:
        r0,r1,s0,s1=r1,r0%r1,s1,s0-r0//r1*s1
    return s0,(r0-s0*x)//y,x*s0+y*(r0-s0*x)//y


## return x,LCM(mods) s.t. x = rem_i (mod_i), x = -1 if such x doesn't exist
## verified by ABC193E
## https://atcoder.jp/contests/abc193/tasks/abc193_e
def crt(rems,mods):
    n=len(rems)
    if n!=len(mods):
        return NotImplemented
    x,d=0,1
    
    for r,m in zip(rems,mods):
        a,b,g=extgcd(d,m)
        x,d=(m*b*x+d*a*r)//g,d*(m//g)
        x%=d

    for r,m in zip(rems,mods):
        if r!=x%m:
            return -1,d

    return x,d

## returns the maximum integer rt s.t. rt*rt<=x
## verified by ABC191D
## https://atcoder.jp/contests/abc191/tasks/abc191_d
def intsqrt(x):
    if x<0:
        return NotImplemented
    rt=int(x**0.5)-1
    while (rt+1)**2<=x:
        rt+=1
    return rt


show_flg=False
show_flg=True

ans=0


n=I()
a=LI()

bi=[]
for i in range(n):
    c=a[i]
    t=0
    while c%2==0:
        c//=2
        t+=1
    bi+=t,
e=2**min(bi)

a=[i//e for i in a]

s=sum(a)
if s%2==1:
    ans=0
else:
    d=[False]*(s+1)
    d[0]=True
    for i in range(n):
        c=a[i]
        for j in range(s,c-1,-1):
            if d[j-c]==True:
                d[j]=True
    if d[s//2]:
        ans=1
        for i in range(n):
            if a[i]%2==1:
                k=i+1
    else:
        ans=0
print(ans)
if ans==1:
    print(k)


exit()
show([i for i,j in enumerate(d) if j],'d')
show(a)